Question

Figure shows a graph in log₁₀ K vs$\frac{1}{T}$where K is rate constant and T is temperature. The straight line BC has slope, | tan q | = $\frac{1}{2.303}$ and an intercept of 5 on y-axis. Thus E_a, the energy of activation is –

• A. 2.303 × 2 cal
• B. 2/2.303 cal
• C. 2 cal
• D. None of these

Answer 1

Answer: (C)

2 cal

Answer 2

log₁₀ K vs $\frac{1}{T}$is linear ;

slope = $\frac{- E_{A}}{2.303R}$ intercept = log₁₀ A

Since, |tan q| = $\frac{1}{2.303}$

\ $\frac{E_{A}}{2.303R}$=$\frac{1}{2.303}$ Ž E_a = R = 2 cal